Tutorials › Statistical Theory I - Probability and Distribution Theory › Counterexamples from Crossing Tails and Density Ratios

Home  /  Tutorials  /  Stochastic Orders and Comparisons

Stochastic Orders and Comparisons · Tutorial 968 of 994

Counterexamples from Crossing Tails and Density Ratios

See how explicit counterexamples expose the limits of moment matching, isolated tail comparisons, and density-ratio checks when establishing distributional order.

Advanced 11 min read

What You'll Learn

  • Define first-order stochastic order through an inequality that must hold at every threshold.
  • Check CDF crossings and likelihood-ratio crossings in finite-support examples.
  • Construct continuous densities with matching low-order moments but crossing CDFs.
  • Verify why matching a single tail probability cannot establish stochastic order.
  • Distinguish evidence about selected summaries from a distribution-wide comparison.

Why One or Two Comparisons Do Not Establish an Order

Distributional orders make claims about an entire distribution, not just its mean, variance, or behavior at a convenient threshold. Two laws can share several moments and still place probability differently across their supports. Their cumulative distribution functions (CDFs) may cross, making neither law uniformly smaller in the usual stochastic sense. Density ratios can also rise and fall rather than satisfy the monotonicity required for likelihood-ratio order.

The aim here is to make those limitations concrete. The examples use finite supports and explicit densities so that the moments, CDFs, tail probabilities, and density ratios can all be checked directly.

Definition: Let \(X\) and \(Y\) have CDFs \(F_X\) and \(F_Y\). We say \(X\) is smaller than \(Y\) in first-order stochastic order, written \(X\le_{\mathrm{st}}Y\), if \(F_X(x)\ge F_Y(x)\) for every real \(x\). Equivalently, \(P(X>x)\le P(Y>x)\) for every real \(x\). If neither inequality holds for all \(x\), the laws are not ordered in this sense. If the CDFs take opposite strict orderings at different points, they cross.

The direction of the CDF inequality can feel counterintuitive: a law shifted toward larger values has a smaller CDF at a given threshold. The equivalent upper-tail statement is often easier to interpret. In either formulation, the condition must hold at every threshold, not just at a few selected ones.

Definition: For laws with positive probability masses \(f_X,f_Y\) on the same ordered discrete support, their likelihood ratio in the direction \(Y\) versus \(X\) is \(f_Y(x)/f_X(x)\). The likelihood-ratio order \(X\le_{\mathrm{lr}}Y\) requires this ratio to be nondecreasing across the support. For positive densities on a common support, the same definition applies to the density ratio. A ratio that is neither nondecreasing nor nonincreasing establishes neither direction of likelihood-ratio order.

Worked Examples on Finite Supports

Worked Example: Matching the Mean and Variance Does Not Order Two Laws

Let \(X\) and \(Y\) take values \(0,1,2,3\), with probability masses, in that order,

$$ f_X=\frac{1}{24}(5,9,3,7), \qquad f_Y=\frac{1}{24}(7,3,9,5). $$

Each vector sums to \(24/24=1\), and every mass is positive. For \(X\), the first two raw moments are

$$ E[X]=\frac{0(5)+1(9)+2(3)+3(7)}{24} =\frac{36}{24}=\frac32, \qquad E[X^2]=\frac{0+9+12+63}{24} =\frac{84}{24}=\frac72. $$

For \(Y\), they are

$$ E[Y]=\frac{0(7)+1(3)+2(9)+3(5)}{24} =\frac{36}{24}=\frac32, \qquad E[Y^2]=\frac{0+3+36+45}{24} =\frac{84}{24}=\frac72. $$

Thus both variances equal \(7/2-(3/2)^2=5/4\). Nevertheless, their CDFs disagree in direction across thresholds. At \(x=0\), \(F_X(0)=5/24<7/24=F_Y(0)\). At \(x=1\), \(F_X(1)=14/24>10/24=F_Y(1)\). At \(x=2\), \(F_X(2)=17/24<19/24=F_Y(2)\). The CDFs cross, so neither law is smaller in first-order stochastic order.

The mass ratio \(f_Y/f_X\) at \(0,1,2,3\) is \(7/5,1/3,3,5/7\). It first decreases, then increases, then decreases, so it is not monotone in either direction. Matching a mean and variance has established neither stochastic nor likelihood-ratio order.

Worked Example: One Matching Tail Probability Does Not Establish Order

Consider laws \(A\) and \(B\) on \(\{0,1,2,3\}\) with masses

$$ f_A=(0.4,0.1,0.1,0.4), \qquad f_B=(0.1,0.4,0.4,0.1). $$

At the threshold \(1\), the upper-tail probabilities agree: \(P(A>1)=0.1+0.4=0.5\) and \(P(B>1)=0.4+0.1=0.5\). But at \(x=0\), \(F_A(0)=0.4>0.1=F_B(0)\), whereas at \(x=2\), \(F_A(2)=0.6<0.9=F_B(2)\). Hence neither CDF inequality holds everywhere, and the laws are not stochastically ordered.

Their likelihood ratio \(f_A/f_B\) is \(4,1/4,1/4,4\), which is also nonmonotone. Equality of one tail probability says only that the CDFs agree at that threshold; it says nothing about which CDF is larger elsewhere.

A Continuous Construction With Matching Moments

Worked Example: Positive Densities With Matching First Two Moments and Crossing CDFs

On \(0\le x\le1\), define \(h(x)=20x^3-30x^2+12x-1\), and set \(p(x)=1+\varepsilon h(x)\) and \(q(x)=1-\varepsilon h(x)\), where \(\varepsilon=1/126\). Outside \([0,1]\), set both densities to zero.

First check that these are densities. On \([0,1]\), the triangle inequality gives \(|h(x)|\le20+30+12+1=63\). Therefore \(p(x),q(x)\ge1-63/126=1/2\). Direct integration gives

$$ \int_0^1h(x)\,dx=5-10+6-1=0, \qquad \int_0^1p(x)\,dx=\int_0^1q(x)\,dx=1. $$

The first two moments agree because \(p-q=2\varepsilon h\) and

$$ \int_0^1 xh(x)\,dx =4-\frac{30}{4}+4-\frac12=0, \qquad \int_0^1 x^2h(x)\,dx =\frac{10}{3}-6+3-\frac13=0. $$

In the first calculation, \(4-30/4+4-1/2=4-7.5+4-0.5=0\). Thus \(E_p[X]=E_q[X]\) and \(E_p[X^2]=E_q[X^2]\); consequently, their variances also agree.

To check the CDFs, integrate their density difference from \(0\) to \(x\):

$$ F_p(x)-F_q(x) =2\varepsilon\int_0^x h(t)\,dt =2\varepsilon\bigl(5x^4-10x^3+6x^2-x\bigr), \qquad 0\le x\le1. $$

The polynomial in parentheses is \(-3/256\) at \(x=1/4\), \(1/16\) at \(x=1/2\), and \(-3/256\) at \(x=3/4\). Since \(2\varepsilon>0\), \(F_p-F_q\) has signs \(-,+,-\) at those points. The CDFs cross, despite the equal first two moments. In addition, the density ratio \(p/q=(1+\varepsilon h)/(1-\varepsilon h)\) is a strictly increasing function of \(h\), but \(h(0)=-1\), \(h(1/4)=7/16\), \(h(3/4)=-7/16\), and \(h(1)=1\). The ratio therefore rises, falls, and rises across these points; it is not monotone in either direction.

What the Comparisons Do and Do Not Tell Us

Theorem: For any real-valued random variables \(X\) and \(Y\), \(X\le_{\mathrm{st}}Y\) if and only if \(P(X>x)\le P(Y>x)\) for every real \(x\).

Proof. By the definition of a CDF, \(P(X>x)=1-F_X(x)\) and \(P(Y>x)=1-F_Y(x)\). Therefore \(F_X(x)\ge F_Y(x)\) is equivalent, by subtracting each side from \(1\), to \(P(X>x)\le P(Y>x)\). Since this equivalence holds separately for every \(x\), it holds for the required all-threshold comparison. \(\square\)

This equivalence explains why a few tail checks are not a substitute for stochastic order: they verify the required inequality only at the checked thresholds. Likewise, matching moments constrains integrals such as \(E[X]\) and \(E[X^2]\), not the complete pattern of CDF values. The continuous construction makes that distinction explicit: two moment constraints can hold exactly while the CDF difference changes sign.

Likelihood-ratio order asks yet another question: whether a density or mass ratio is monotone over the common support. A nonmonotone ratio rules out both directions of that order in the examples above, but it should not be confused with a CDF crossing. These are different comparisons; one concerns a ratio of local probabilities or densities, the other cumulative probability up to each threshold.

Common Mistakes

  • Checking only one threshold. Equal tail probabilities at one point give equality there, not an ordering at all other points. Compare the full CDFs or establish a general condition that guarantees the required inequality everywhere.
  • Treating equal means and variances as distributional equality. These are only two summaries. The discrete and continuous examples have matching moments of these orders but distinct, crossing CDFs.
  • Calling a ratio monotone because it crosses one once. Crossing the level one does not imply a ratio increases or decreases throughout its support. Check the ratio's behavior across the support, not just its intersections with one.
  • Reversing the CDF direction. For \(X\le_{\mathrm{st}}Y\), the required direction is \(F_X(x)\ge F_Y(x)\), equivalently \(P(X>x)\le P(Y>x)\), for every threshold.
Key takeaway: A mean, variance, selected tail probabilities, or a few density-ratio values cannot establish a distributional order by themselves. Stochastic order requires a consistent CDF (or upper-tail) inequality at every threshold; likelihood-ratio order requires a monotone ratio across the common support.

Check Your Understanding

Use the examples and definitions above to answer these questions.

  1. For first-order stochastic order, which way must the CDF inequality point for \(X\le_{\mathrm{st}}Y\), and why?
  2. In the first discrete example, verify \(F_X(1)\) and \(F_Y(1)\) from the masses. What does their comparison imply together with the comparison at \(x=0\)?
  3. Why do matching first and second moments in the continuous example imply matching variances?
  4. Two laws have equal upper-tail probabilities at \(x=4\). What does this establish about their CDFs at \(4\), and what does it leave unresolved?
  5. What feature of a density ratio is required for likelihood-ratio order, and why is merely crossing the value one insufficient?

Previous

Next

Back to Tutorials